The results of that equation, while adequate for most manual machining operations, is often inadequate for high spindle horsepower N/C machines. In addition to being uneconomical, a feedrate that is too slow can hasten cutting tool failure. As each tooth peels off a chip, it also work hardens the workpiece surface. The depth of this workhardened layer can range from a few millionths of an inch to over a thousandth of an inch. If the feedrate is too slow, the cutting edge of each tooth can be cutting through the workhardened layer (dulling the cutting edge) instead of slicing beneath the layer. A feedrate that is too high can stall the spindle or break the cutter--or both. The feedrate that is required to stall the spindle depends upon the following factors: Horsepower of the spindle drive motor, Material Machining Factor (MMF), the machineability of the workpiece material, Material Removal Rate (MRR), the volume (cubic inches) of material being removed per minute. MRR, in turn, is dependent on the depth of cut, diameter of the cutter, and feedrate. Geometry of the cutter, Sharpness of the cutter. The geometry of the cutter is a minor factor and can be ignored. Good machining practice presupposes the cutter is sharp, so that factor can be ignored. Material Removal Rate (MRR) For milling, the material removal rate (MRR) can be calculated as follows: MMR = depth of cut * the width of cut * spindle speed * the feed per tooth (chip load) * the number of teeth * the spindle speed (RPM). For lathe cuts, the MRR is the product of the mean workpiece circumference * the depth of cut * feedrate (in inches per revolution) * the spindle speed. The mean workpiece circumference is the circumference halfway down the depth of the cut. For drilling, the MRR is the product of the area of a circle the diameter of the drill (drill diameter squared * pi/4) * the feedrate in inches per revolution * the spindle speed in RPM. The horsepower required to run the spindle is the product of the Material Removal Rate (MRR) * the Material Machining Factor (MMF). The MMF, a function of the machineability rating of the workpiece material, is shown in Table 3.3. Table 3.3 Material Machining Factors (MMF) Material MMF Aluminum0.12 Brass0.4 Low-carbon steel1.25 High-carbon & low alloy steel1.5 - 1.9 Stainless and tool steels1.8 - 2.2 Superalloys2.5 - 5.0 More Mathematics For example, suppose a 1-inch-diameter end mill was making a 1-inch-deep cut in low-carbon steel with a feedrate of 10 inches per minute (IPM). The MRR is 1 (depth) * 1 (width) * 10 (IPM) = 10. Table 3.3 shows low-carbon steel has an MMF of 1.25. Thus, it would require 10 (MRR) * 1.25 (MMF) = 12.5 HP to drive the spindle. If the spindle was driven with a 5 HP motor, it could stall and perhaps break the cutter. The feedrate (or depth of cut) would have to be reduced to lower the MRR, so a 5 HP or lower load was placed on the spindle. Since: HP = MRR * MMF and MRR = depth of cut * width of cut * feedrate it follows that: HP = MMF * depth of cut * width of cut * feedrate. Thus, Feedrate = HP --------------------------------------- MMF * depth of cut * width of cut In the previous example, if the spindle motor were 5 HP Feedrate = 5 HP --------------------------------------- 1.25 (MMF) * 1 (depth) * 1 (width) = 4 IPM   Per tooth chip load = feedrate (IPM) --------------------------------------- spindle RPM * cutter teeth Assuming a spindle speed of 400 RPM and a 2-flute cutter, Chip load = 4.0 IPM --------------------------------------- 400 RPM * 2 teeth = 0.005 inch per tooth Suppose in the previous example the workpiece material is changed to aluminum (with an MMF of 0.12). Spindle horsepower is 5. Spindle RPM is 4000. Feedrate = 5 HP --------------------------------------- 0.12 (MMF) * 1 (depth) * 1 (width) = 41.7 IPM   Chip load = 41.7 IPM --------------------------------------- 4000 RPM * 2 teeth = 0.0052 inch per tooth Next:  Review Questions for Chapter 3 Back to Contents Page Updated Jan. 9, 2002 Copyright © 1988-2002 by George Stanton and Bill Hemphill All Rights Reserved

That's a good question. You'll want to drill only as deep as you'll need to for the screw to fit. Wood is a bit different in that you can ...

Except for those rare few souls that write N/C programs for the sheer joy of it, the ultimate purpose of owning, programming, or operating an N/C machine tool is to make a buck. The buck is made by performing a service or making a product better and at lower cost than the competetion can do it. One of the major factors in machine tool economics is time. The longer it takes to do something, the more it costs. The length of time it takes to make a piecepart using N/C is a direct function of the feedrate used. The slower the feedrate, the longer it will take to run the piecepart. Other factors, such as making unnecessary cutting passes, are also important, but using the correct feedrate (and spindle speed) is perhaps the most important factor in the economics of N/C. The optimum feedrate is one that maximizes the Material Removal Rate (MRR) without stalling the spindle or breaking the cutter. Determining the Ideal Feedrate The classic method of determining the feedrate is to decide upon a chip load (how thick a chip you wish each tooth to peel off), typically from 0.003 to 0.020 inch. Then calculate what the spindle speed should be. Next, count the number of teeth on the cutter. Then multiply the three together: Feedrate (IPM) = chipload (IN) * RPM * teeth The results of that equation, while adequate for most manual machining operations, is often inadequate for high spindle horsepower N/C machines. In addition to being uneconomical, a feedrate that is too slow can hasten cutting tool failure. As each tooth peels off a chip, it also work hardens the workpiece surface. The depth of this workhardened layer can range from a few millionths of an inch to over a thousandth of an inch. If the feedrate is too slow, the cutting edge of each tooth can be cutting through the workhardened layer (dulling the cutting edge) instead of slicing beneath the layer. A feedrate that is too high can stall the spindle or break the cutter--or both. The feedrate that is required to stall the spindle depends upon the following factors: Horsepower of the spindle drive motor, Material Machining Factor (MMF), the machineability of the workpiece material, Material Removal Rate (MRR), the volume (cubic inches) of material being removed per minute. MRR, in turn, is dependent on the depth of cut, diameter of the cutter, and feedrate. Geometry of the cutter, Sharpness of the cutter. The geometry of the cutter is a minor factor and can be ignored. Good machining practice presupposes the cutter is sharp, so that factor can be ignored. Material Removal Rate (MRR) For milling, the material removal rate (MRR) can be calculated as follows: MMR = depth of cut * the width of cut * spindle speed * the feed per tooth (chip load) * the number of teeth * the spindle speed (RPM). For lathe cuts, the MRR is the product of the mean workpiece circumference * the depth of cut * feedrate (in inches per revolution) * the spindle speed. The mean workpiece circumference is the circumference halfway down the depth of the cut. For drilling, the MRR is the product of the area of a circle the diameter of the drill (drill diameter squared * pi/4) * the feedrate in inches per revolution * the spindle speed in RPM. The horsepower required to run the spindle is the product of the Material Removal Rate (MRR) * the Material Machining Factor (MMF). The MMF, a function of the machineability rating of the workpiece material, is shown in Table 3.3. Table 3.3 Material Machining Factors (MMF) Material MMF Aluminum0.12 Brass0.4 Low-carbon steel1.25 High-carbon & low alloy steel1.5 - 1.9 Stainless and tool steels1.8 - 2.2 Superalloys2.5 - 5.0 More Mathematics For example, suppose a 1-inch-diameter end mill was making a 1-inch-deep cut in low-carbon steel with a feedrate of 10 inches per minute (IPM). The MRR is 1 (depth) * 1 (width) * 10 (IPM) = 10. Table 3.3 shows low-carbon steel has an MMF of 1.25. Thus, it would require 10 (MRR) * 1.25 (MMF) = 12.5 HP to drive the spindle. If the spindle was driven with a 5 HP motor, it could stall and perhaps break the cutter. The feedrate (or depth of cut) would have to be reduced to lower the MRR, so a 5 HP or lower load was placed on the spindle. Since: HP = MRR * MMF and MRR = depth of cut * width of cut * feedrate it follows that: HP = MMF * depth of cut * width of cut * feedrate. Thus, Feedrate = HP --------------------------------------- MMF * depth of cut * width of cut In the previous example, if the spindle motor were 5 HP Feedrate = 5 HP --------------------------------------- 1.25 (MMF) * 1 (depth) * 1 (width) = 4 IPM   Per tooth chip load = feedrate (IPM) --------------------------------------- spindle RPM * cutter teeth Assuming a spindle speed of 400 RPM and a 2-flute cutter, Chip load = 4.0 IPM --------------------------------------- 400 RPM * 2 teeth = 0.005 inch per tooth Suppose in the previous example the workpiece material is changed to aluminum (with an MMF of 0.12). Spindle horsepower is 5. Spindle RPM is 4000. Feedrate = 5 HP --------------------------------------- 0.12 (MMF) * 1 (depth) * 1 (width) = 41.7 IPM   Chip load = 41.7 IPM --------------------------------------- 4000 RPM * 2 teeth = 0.0052 inch per tooth Next:  Review Questions for Chapter 3 Back to Contents Page Updated Jan. 9, 2002 Copyright © 1988-2002 by George Stanton and Bill Hemphill All Rights Reserved

2021414 — Strain hardening, also referred to as cold working or work hardening, is the process where metals are made stronger and harder through permanent deformation.

... End HSSCo Roughing Center Cutting End Mill-TiN Series/List #4613G. HSS-PM. Item # 44987. UNSPSC: 27112803. Brand: Morse Cutting Tools. MFG#: 44987. UPC ...

and MRR = depth of cut * width of cut * feedrate it follows that: HP = MMF * depth of cut * width of cut * feedrate. Thus, Feedrate = HP --------------------------------------- MMF * depth of cut * width of cut In the previous example, if the spindle motor were 5 HP Feedrate = 5 HP --------------------------------------- 1.25 (MMF) * 1 (depth) * 1 (width) = 4 IPM   Per tooth chip load = feedrate (IPM) --------------------------------------- spindle RPM * cutter teeth Assuming a spindle speed of 400 RPM and a 2-flute cutter, Chip load = 4.0 IPM --------------------------------------- 400 RPM * 2 teeth = 0.005 inch per tooth Suppose in the previous example the workpiece material is changed to aluminum (with an MMF of 0.12). Spindle horsepower is 5. Spindle RPM is 4000. Feedrate = 5 HP --------------------------------------- 0.12 (MMF) * 1 (depth) * 1 (width) = 41.7 IPM   Chip load = 41.7 IPM --------------------------------------- 4000 RPM * 2 teeth = 0.0052 inch per tooth Next:  Review Questions for Chapter 3 Back to Contents Page Updated Jan. 9, 2002 Copyright © 1988-2002 by George Stanton and Bill Hemphill All Rights Reserved

The geometry of the cutter is a minor factor and can be ignored. Good machining practice presupposes the cutter is sharp, so that factor can be ignored. Material Removal Rate (MRR) For milling, the material removal rate (MRR) can be calculated as follows: MMR = depth of cut * the width of cut * spindle speed * the feed per tooth (chip load) * the number of teeth * the spindle speed (RPM). For lathe cuts, the MRR is the product of the mean workpiece circumference * the depth of cut * feedrate (in inches per revolution) * the spindle speed. The mean workpiece circumference is the circumference halfway down the depth of the cut. For drilling, the MRR is the product of the area of a circle the diameter of the drill (drill diameter squared * pi/4) * the feedrate in inches per revolution * the spindle speed in RPM. The horsepower required to run the spindle is the product of the Material Removal Rate (MRR) * the Material Machining Factor (MMF). The MMF, a function of the machineability rating of the workpiece material, is shown in Table 3.3. Table 3.3 Material Machining Factors (MMF) Material MMF Aluminum0.12 Brass0.4 Low-carbon steel1.25 High-carbon & low alloy steel1.5 - 1.9 Stainless and tool steels1.8 - 2.2 Superalloys2.5 - 5.0 More Mathematics For example, suppose a 1-inch-diameter end mill was making a 1-inch-deep cut in low-carbon steel with a feedrate of 10 inches per minute (IPM). The MRR is 1 (depth) * 1 (width) * 10 (IPM) = 10. Table 3.3 shows low-carbon steel has an MMF of 1.25. Thus, it would require 10 (MRR) * 1.25 (MMF) = 12.5 HP to drive the spindle. If the spindle was driven with a 5 HP motor, it could stall and perhaps break the cutter. The feedrate (or depth of cut) would have to be reduced to lower the MRR, so a 5 HP or lower load was placed on the spindle. Since: HP = MRR * MMF and MRR = depth of cut * width of cut * feedrate it follows that: HP = MMF * depth of cut * width of cut * feedrate. Thus, Feedrate = HP --------------------------------------- MMF * depth of cut * width of cut In the previous example, if the spindle motor were 5 HP Feedrate = 5 HP --------------------------------------- 1.25 (MMF) * 1 (depth) * 1 (width) = 4 IPM   Per tooth chip load = feedrate (IPM) --------------------------------------- spindle RPM * cutter teeth Assuming a spindle speed of 400 RPM and a 2-flute cutter, Chip load = 4.0 IPM --------------------------------------- 400 RPM * 2 teeth = 0.005 inch per tooth Suppose in the previous example the workpiece material is changed to aluminum (with an MMF of 0.12). Spindle horsepower is 5. Spindle RPM is 4000. Feedrate = 5 HP --------------------------------------- 0.12 (MMF) * 1 (depth) * 1 (width) = 41.7 IPM   Chip load = 41.7 IPM --------------------------------------- 4000 RPM * 2 teeth = 0.0052 inch per tooth Next:  Review Questions for Chapter 3 Back to Contents Page Updated Jan. 9, 2002 Copyright © 1988-2002 by George Stanton and Bill Hemphill All Rights Reserved

A feedrate that is too high can stall the spindle or break the cutter--or both. The feedrate that is required to stall the spindle depends upon the following factors: Horsepower of the spindle drive motor, Material Machining Factor (MMF), the machineability of the workpiece material, Material Removal Rate (MRR), the volume (cubic inches) of material being removed per minute. MRR, in turn, is dependent on the depth of cut, diameter of the cutter, and feedrate. Geometry of the cutter, Sharpness of the cutter. The geometry of the cutter is a minor factor and can be ignored. Good machining practice presupposes the cutter is sharp, so that factor can be ignored. Material Removal Rate (MRR) For milling, the material removal rate (MRR) can be calculated as follows: MMR = depth of cut * the width of cut * spindle speed * the feed per tooth (chip load) * the number of teeth * the spindle speed (RPM). For lathe cuts, the MRR is the product of the mean workpiece circumference * the depth of cut * feedrate (in inches per revolution) * the spindle speed. The mean workpiece circumference is the circumference halfway down the depth of the cut. For drilling, the MRR is the product of the area of a circle the diameter of the drill (drill diameter squared * pi/4) * the feedrate in inches per revolution * the spindle speed in RPM. The horsepower required to run the spindle is the product of the Material Removal Rate (MRR) * the Material Machining Factor (MMF). The MMF, a function of the machineability rating of the workpiece material, is shown in Table 3.3. Table 3.3 Material Machining Factors (MMF) Material MMF Aluminum0.12 Brass0.4 Low-carbon steel1.25 High-carbon & low alloy steel1.5 - 1.9 Stainless and tool steels1.8 - 2.2 Superalloys2.5 - 5.0 More Mathematics For example, suppose a 1-inch-diameter end mill was making a 1-inch-deep cut in low-carbon steel with a feedrate of 10 inches per minute (IPM). The MRR is 1 (depth) * 1 (width) * 10 (IPM) = 10. Table 3.3 shows low-carbon steel has an MMF of 1.25. Thus, it would require 10 (MRR) * 1.25 (MMF) = 12.5 HP to drive the spindle. If the spindle was driven with a 5 HP motor, it could stall and perhaps break the cutter. The feedrate (or depth of cut) would have to be reduced to lower the MRR, so a 5 HP or lower load was placed on the spindle. Since: HP = MRR * MMF and MRR = depth of cut * width of cut * feedrate it follows that: HP = MMF * depth of cut * width of cut * feedrate. Thus, Feedrate = HP --------------------------------------- MMF * depth of cut * width of cut In the previous example, if the spindle motor were 5 HP Feedrate = 5 HP --------------------------------------- 1.25 (MMF) * 1 (depth) * 1 (width) = 4 IPM   Per tooth chip load = feedrate (IPM) --------------------------------------- spindle RPM * cutter teeth Assuming a spindle speed of 400 RPM and a 2-flute cutter, Chip load = 4.0 IPM --------------------------------------- 400 RPM * 2 teeth = 0.005 inch per tooth Suppose in the previous example the workpiece material is changed to aluminum (with an MMF of 0.12). Spindle horsepower is 5. Spindle RPM is 4000. Feedrate = 5 HP --------------------------------------- 0.12 (MMF) * 1 (depth) * 1 (width) = 41.7 IPM   Chip load = 41.7 IPM --------------------------------------- 4000 RPM * 2 teeth = 0.0052 inch per tooth Next:  Review Questions for Chapter 3 Back to Contents Page Updated Jan. 9, 2002 Copyright © 1988-2002 by George Stanton and Bill Hemphill All Rights Reserved

For lathe cuts, the MRR is the product of the mean workpiece circumference * the depth of cut * feedrate (in inches per revolution) * the spindle speed. The mean workpiece circumference is the circumference halfway down the depth of the cut. For drilling, the MRR is the product of the area of a circle the diameter of the drill (drill diameter squared * pi/4) * the feedrate in inches per revolution * the spindle speed in RPM. The horsepower required to run the spindle is the product of the Material Removal Rate (MRR) * the Material Machining Factor (MMF). The MMF, a function of the machineability rating of the workpiece material, is shown in Table 3.3. Table 3.3 Material Machining Factors (MMF) Material MMF Aluminum0.12 Brass0.4 Low-carbon steel1.25 High-carbon & low alloy steel1.5 - 1.9 Stainless and tool steels1.8 - 2.2 Superalloys2.5 - 5.0 More Mathematics For example, suppose a 1-inch-diameter end mill was making a 1-inch-deep cut in low-carbon steel with a feedrate of 10 inches per minute (IPM). The MRR is 1 (depth) * 1 (width) * 10 (IPM) = 10. Table 3.3 shows low-carbon steel has an MMF of 1.25. Thus, it would require 10 (MRR) * 1.25 (MMF) = 12.5 HP to drive the spindle. If the spindle was driven with a 5 HP motor, it could stall and perhaps break the cutter. The feedrate (or depth of cut) would have to be reduced to lower the MRR, so a 5 HP or lower load was placed on the spindle. Since: HP = MRR * MMF and MRR = depth of cut * width of cut * feedrate it follows that: HP = MMF * depth of cut * width of cut * feedrate. Thus, Feedrate = HP --------------------------------------- MMF * depth of cut * width of cut In the previous example, if the spindle motor were 5 HP Feedrate = 5 HP --------------------------------------- 1.25 (MMF) * 1 (depth) * 1 (width) = 4 IPM   Per tooth chip load = feedrate (IPM) --------------------------------------- spindle RPM * cutter teeth Assuming a spindle speed of 400 RPM and a 2-flute cutter, Chip load = 4.0 IPM --------------------------------------- 400 RPM * 2 teeth = 0.005 inch per tooth Suppose in the previous example the workpiece material is changed to aluminum (with an MMF of 0.12). Spindle horsepower is 5. Spindle RPM is 4000. Feedrate = 5 HP --------------------------------------- 0.12 (MMF) * 1 (depth) * 1 (width) = 41.7 IPM   Chip load = 41.7 IPM --------------------------------------- 4000 RPM * 2 teeth = 0.0052 inch per tooth Next:  Review Questions for Chapter 3 Back to Contents Page Updated Jan. 9, 2002 Copyright © 1988-2002 by George Stanton and Bill Hemphill All Rights Reserved

Since: HP = MRR * MMF and MRR = depth of cut * width of cut * feedrate it follows that: HP = MMF * depth of cut * width of cut * feedrate. Thus, Feedrate = HP --------------------------------------- MMF * depth of cut * width of cut In the previous example, if the spindle motor were 5 HP Feedrate = 5 HP --------------------------------------- 1.25 (MMF) * 1 (depth) * 1 (width) = 4 IPM   Per tooth chip load = feedrate (IPM) --------------------------------------- spindle RPM * cutter teeth Assuming a spindle speed of 400 RPM and a 2-flute cutter, Chip load = 4.0 IPM --------------------------------------- 400 RPM * 2 teeth = 0.005 inch per tooth Suppose in the previous example the workpiece material is changed to aluminum (with an MMF of 0.12). Spindle horsepower is 5. Spindle RPM is 4000. Feedrate = 5 HP --------------------------------------- 0.12 (MMF) * 1 (depth) * 1 (width) = 41.7 IPM   Chip load = 41.7 IPM --------------------------------------- 4000 RPM * 2 teeth = 0.0052 inch per tooth Next:  Review Questions for Chapter 3 Back to Contents Page Updated Jan. 9, 2002 Copyright © 1988-2002 by George Stanton and Bill Hemphill All Rights Reserved

Dec 13, 2020 — Well, that's basically what any screw does, that is threaded into the plastic the usual way. A thread forming tool has the slight benefit, that ...

For example, suppose a 1-inch-diameter end mill was making a 1-inch-deep cut in low-carbon steel with a feedrate of 10 inches per minute (IPM). The MRR is 1 (depth) * 1 (width) * 10 (IPM) = 10. Table 3.3 shows low-carbon steel has an MMF of 1.25. Thus, it would require 10 (MRR) * 1.25 (MMF) = 12.5 HP to drive the spindle. If the spindle was driven with a 5 HP motor, it could stall and perhaps break the cutter. The feedrate (or depth of cut) would have to be reduced to lower the MRR, so a 5 HP or lower load was placed on the spindle. Since: HP = MRR * MMF and MRR = depth of cut * width of cut * feedrate it follows that: HP = MMF * depth of cut * width of cut * feedrate. Thus, Feedrate = HP --------------------------------------- MMF * depth of cut * width of cut In the previous example, if the spindle motor were 5 HP Feedrate = 5 HP --------------------------------------- 1.25 (MMF) * 1 (depth) * 1 (width) = 4 IPM   Per tooth chip load = feedrate (IPM) --------------------------------------- spindle RPM * cutter teeth Assuming a spindle speed of 400 RPM and a 2-flute cutter, Chip load = 4.0 IPM --------------------------------------- 400 RPM * 2 teeth = 0.005 inch per tooth Suppose in the previous example the workpiece material is changed to aluminum (with an MMF of 0.12). Spindle horsepower is 5. Spindle RPM is 4000. Feedrate = 5 HP --------------------------------------- 0.12 (MMF) * 1 (depth) * 1 (width) = 41.7 IPM   Chip load = 41.7 IPM --------------------------------------- 4000 RPM * 2 teeth = 0.0052 inch per tooth Next:  Review Questions for Chapter 3 Back to Contents Page Updated Jan. 9, 2002 Copyright © 1988-2002 by George Stanton and Bill Hemphill All Rights Reserved

HP = MMF * depth of cut * width of cut * feedrate. Thus, Feedrate = HP --------------------------------------- MMF * depth of cut * width of cut In the previous example, if the spindle motor were 5 HP Feedrate = 5 HP --------------------------------------- 1.25 (MMF) * 1 (depth) * 1 (width) = 4 IPM   Per tooth chip load = feedrate (IPM) --------------------------------------- spindle RPM * cutter teeth Assuming a spindle speed of 400 RPM and a 2-flute cutter, Chip load = 4.0 IPM --------------------------------------- 400 RPM * 2 teeth = 0.005 inch per tooth Suppose in the previous example the workpiece material is changed to aluminum (with an MMF of 0.12). Spindle horsepower is 5. Spindle RPM is 4000. Feedrate = 5 HP --------------------------------------- 0.12 (MMF) * 1 (depth) * 1 (width) = 41.7 IPM   Chip load = 41.7 IPM --------------------------------------- 4000 RPM * 2 teeth = 0.0052 inch per tooth Next:  Review Questions for Chapter 3 Back to Contents Page Updated Jan. 9, 2002 Copyright © 1988-2002 by George Stanton and Bill Hemphill All Rights Reserved

Sep 4, 2023 — a mechanical system used for coupling a cylindrical shaft to other cylindrical part. There are several types of key shapes: rectangular, rounded, round, semi- ...

202313 — Guy Harvey, Inc. and Tennessee-based Groove Life have announced the addition of four new Silicone Ring styles and two new Apple Watch bands ...

202215 — Using the right shape of an insert with turning applications is essential, and it is determined by the appropriate point angle for strength and ...

In addition to being uneconomical, a feedrate that is too slow can hasten cutting tool failure. As each tooth peels off a chip, it also work hardens the workpiece surface. The depth of this workhardened layer can range from a few millionths of an inch to over a thousandth of an inch. If the feedrate is too slow, the cutting edge of each tooth can be cutting through the workhardened layer (dulling the cutting edge) instead of slicing beneath the layer. A feedrate that is too high can stall the spindle or break the cutter--or both. The feedrate that is required to stall the spindle depends upon the following factors: Horsepower of the spindle drive motor, Material Machining Factor (MMF), the machineability of the workpiece material, Material Removal Rate (MRR), the volume (cubic inches) of material being removed per minute. MRR, in turn, is dependent on the depth of cut, diameter of the cutter, and feedrate. Geometry of the cutter, Sharpness of the cutter. The geometry of the cutter is a minor factor and can be ignored. Good machining practice presupposes the cutter is sharp, so that factor can be ignored. Material Removal Rate (MRR) For milling, the material removal rate (MRR) can be calculated as follows: MMR = depth of cut * the width of cut * spindle speed * the feed per tooth (chip load) * the number of teeth * the spindle speed (RPM). For lathe cuts, the MRR is the product of the mean workpiece circumference * the depth of cut * feedrate (in inches per revolution) * the spindle speed. The mean workpiece circumference is the circumference halfway down the depth of the cut. For drilling, the MRR is the product of the area of a circle the diameter of the drill (drill diameter squared * pi/4) * the feedrate in inches per revolution * the spindle speed in RPM. The horsepower required to run the spindle is the product of the Material Removal Rate (MRR) * the Material Machining Factor (MMF). The MMF, a function of the machineability rating of the workpiece material, is shown in Table 3.3. Table 3.3 Material Machining Factors (MMF) Material MMF Aluminum0.12 Brass0.4 Low-carbon steel1.25 High-carbon & low alloy steel1.5 - 1.9 Stainless and tool steels1.8 - 2.2 Superalloys2.5 - 5.0 More Mathematics For example, suppose a 1-inch-diameter end mill was making a 1-inch-deep cut in low-carbon steel with a feedrate of 10 inches per minute (IPM). The MRR is 1 (depth) * 1 (width) * 10 (IPM) = 10. Table 3.3 shows low-carbon steel has an MMF of 1.25. Thus, it would require 10 (MRR) * 1.25 (MMF) = 12.5 HP to drive the spindle. If the spindle was driven with a 5 HP motor, it could stall and perhaps break the cutter. The feedrate (or depth of cut) would have to be reduced to lower the MRR, so a 5 HP or lower load was placed on the spindle. Since: HP = MRR * MMF and MRR = depth of cut * width of cut * feedrate it follows that: HP = MMF * depth of cut * width of cut * feedrate. Thus, Feedrate = HP --------------------------------------- MMF * depth of cut * width of cut In the previous example, if the spindle motor were 5 HP Feedrate = 5 HP --------------------------------------- 1.25 (MMF) * 1 (depth) * 1 (width) = 4 IPM   Per tooth chip load = feedrate (IPM) --------------------------------------- spindle RPM * cutter teeth Assuming a spindle speed of 400 RPM and a 2-flute cutter, Chip load = 4.0 IPM --------------------------------------- 400 RPM * 2 teeth = 0.005 inch per tooth Suppose in the previous example the workpiece material is changed to aluminum (with an MMF of 0.12). Spindle horsepower is 5. Spindle RPM is 4000. Feedrate = 5 HP --------------------------------------- 0.12 (MMF) * 1 (depth) * 1 (width) = 41.7 IPM   Chip load = 41.7 IPM --------------------------------------- 4000 RPM * 2 teeth = 0.0052 inch per tooth Next:  Review Questions for Chapter 3 Back to Contents Page Updated Jan. 9, 2002 Copyright © 1988-2002 by George Stanton and Bill Hemphill All Rights Reserved

MRR = depth of cut * width of cut * feedrate it follows that: HP = MMF * depth of cut * width of cut * feedrate. Thus, Feedrate = HP --------------------------------------- MMF * depth of cut * width of cut In the previous example, if the spindle motor were 5 HP Feedrate = 5 HP --------------------------------------- 1.25 (MMF) * 1 (depth) * 1 (width) = 4 IPM   Per tooth chip load = feedrate (IPM) --------------------------------------- spindle RPM * cutter teeth Assuming a spindle speed of 400 RPM and a 2-flute cutter, Chip load = 4.0 IPM --------------------------------------- 400 RPM * 2 teeth = 0.005 inch per tooth Suppose in the previous example the workpiece material is changed to aluminum (with an MMF of 0.12). Spindle horsepower is 5. Spindle RPM is 4000. Feedrate = 5 HP --------------------------------------- 0.12 (MMF) * 1 (depth) * 1 (width) = 41.7 IPM   Chip load = 41.7 IPM --------------------------------------- 4000 RPM * 2 teeth = 0.0052 inch per tooth Next:  Review Questions for Chapter 3 Back to Contents Page Updated Jan. 9, 2002 Copyright © 1988-2002 by George Stanton and Bill Hemphill All Rights Reserved

HP = MRR * MMF and MRR = depth of cut * width of cut * feedrate it follows that: HP = MMF * depth of cut * width of cut * feedrate. Thus, Feedrate = HP --------------------------------------- MMF * depth of cut * width of cut In the previous example, if the spindle motor were 5 HP Feedrate = 5 HP --------------------------------------- 1.25 (MMF) * 1 (depth) * 1 (width) = 4 IPM   Per tooth chip load = feedrate (IPM) --------------------------------------- spindle RPM * cutter teeth Assuming a spindle speed of 400 RPM and a 2-flute cutter, Chip load = 4.0 IPM --------------------------------------- 400 RPM * 2 teeth = 0.005 inch per tooth Suppose in the previous example the workpiece material is changed to aluminum (with an MMF of 0.12). Spindle horsepower is 5. Spindle RPM is 4000. Feedrate = 5 HP --------------------------------------- 0.12 (MMF) * 1 (depth) * 1 (width) = 41.7 IPM   Chip load = 41.7 IPM --------------------------------------- 4000 RPM * 2 teeth = 0.0052 inch per tooth Next:  Review Questions for Chapter 3 Back to Contents Page Updated Jan. 9, 2002 Copyright © 1988-2002 by George Stanton and Bill Hemphill All Rights Reserved

Chip load = 41.7 IPM --------------------------------------- 4000 RPM * 2 teeth = 0.0052 inch per tooth Next:  Review Questions for Chapter 3 Back to Contents Page Updated Jan. 9, 2002 Copyright © 1988-2002 by George Stanton and Bill Hemphill All Rights Reserved

We are capable of building chip conveyors and material handling equipment of just about any size. In addition, multiple conveyor belt options are offered to meet a variety of applications. Above all, Jorgensen is a one stop shop for all of your industrial conveyor and coolant filtration needs.

The Klein Tools 1-Inch x 54-Inch Flex Bit Auger works well either horizontally or vertically when you need to drill within a wall.

Suppose in the previous example the workpiece material is changed to aluminum (with an MMF of 0.12). Spindle horsepower is 5. Spindle RPM is 4000.

The primary criterion for consideration of papers in Thin-Walled Structures is that they must be concerned with thin–walled structures or the basic problems ...

One of the major factors in machine tool economics is time. The longer it takes to do something, the more it costs. The length of time it takes to make a piecepart using N/C is a direct function of the feedrate used. The slower the feedrate, the longer it will take to run the piecepart. Other factors, such as making unnecessary cutting passes, are also important, but using the correct feedrate (and spindle speed) is perhaps the most important factor in the economics of N/C. The optimum feedrate is one that maximizes the Material Removal Rate (MRR) without stalling the spindle or breaking the cutter. Determining the Ideal Feedrate The classic method of determining the feedrate is to decide upon a chip load (how thick a chip you wish each tooth to peel off), typically from 0.003 to 0.020 inch. Then calculate what the spindle speed should be. Next, count the number of teeth on the cutter. Then multiply the three together: Feedrate (IPM) = chipload (IN) * RPM * teeth The results of that equation, while adequate for most manual machining operations, is often inadequate for high spindle horsepower N/C machines. In addition to being uneconomical, a feedrate that is too slow can hasten cutting tool failure. As each tooth peels off a chip, it also work hardens the workpiece surface. The depth of this workhardened layer can range from a few millionths of an inch to over a thousandth of an inch. If the feedrate is too slow, the cutting edge of each tooth can be cutting through the workhardened layer (dulling the cutting edge) instead of slicing beneath the layer. A feedrate that is too high can stall the spindle or break the cutter--or both. The feedrate that is required to stall the spindle depends upon the following factors: Horsepower of the spindle drive motor, Material Machining Factor (MMF), the machineability of the workpiece material, Material Removal Rate (MRR), the volume (cubic inches) of material being removed per minute. MRR, in turn, is dependent on the depth of cut, diameter of the cutter, and feedrate. Geometry of the cutter, Sharpness of the cutter. The geometry of the cutter is a minor factor and can be ignored. Good machining practice presupposes the cutter is sharp, so that factor can be ignored. Material Removal Rate (MRR) For milling, the material removal rate (MRR) can be calculated as follows: MMR = depth of cut * the width of cut * spindle speed * the feed per tooth (chip load) * the number of teeth * the spindle speed (RPM). For lathe cuts, the MRR is the product of the mean workpiece circumference * the depth of cut * feedrate (in inches per revolution) * the spindle speed. The mean workpiece circumference is the circumference halfway down the depth of the cut. For drilling, the MRR is the product of the area of a circle the diameter of the drill (drill diameter squared * pi/4) * the feedrate in inches per revolution * the spindle speed in RPM. The horsepower required to run the spindle is the product of the Material Removal Rate (MRR) * the Material Machining Factor (MMF). The MMF, a function of the machineability rating of the workpiece material, is shown in Table 3.3. Table 3.3 Material Machining Factors (MMF) Material MMF Aluminum0.12 Brass0.4 Low-carbon steel1.25 High-carbon & low alloy steel1.5 - 1.9 Stainless and tool steels1.8 - 2.2 Superalloys2.5 - 5.0 More Mathematics For example, suppose a 1-inch-diameter end mill was making a 1-inch-deep cut in low-carbon steel with a feedrate of 10 inches per minute (IPM). The MRR is 1 (depth) * 1 (width) * 10 (IPM) = 10. Table 3.3 shows low-carbon steel has an MMF of 1.25. Thus, it would require 10 (MRR) * 1.25 (MMF) = 12.5 HP to drive the spindle. If the spindle was driven with a 5 HP motor, it could stall and perhaps break the cutter. The feedrate (or depth of cut) would have to be reduced to lower the MRR, so a 5 HP or lower load was placed on the spindle. Since: HP = MRR * MMF and MRR = depth of cut * width of cut * feedrate it follows that: HP = MMF * depth of cut * width of cut * feedrate. Thus, Feedrate = HP --------------------------------------- MMF * depth of cut * width of cut In the previous example, if the spindle motor were 5 HP Feedrate = 5 HP --------------------------------------- 1.25 (MMF) * 1 (depth) * 1 (width) = 4 IPM   Per tooth chip load = feedrate (IPM) --------------------------------------- spindle RPM * cutter teeth Assuming a spindle speed of 400 RPM and a 2-flute cutter, Chip load = 4.0 IPM --------------------------------------- 400 RPM * 2 teeth = 0.005 inch per tooth Suppose in the previous example the workpiece material is changed to aluminum (with an MMF of 0.12). Spindle horsepower is 5. Spindle RPM is 4000. Feedrate = 5 HP --------------------------------------- 0.12 (MMF) * 1 (depth) * 1 (width) = 41.7 IPM   Chip load = 41.7 IPM --------------------------------------- 4000 RPM * 2 teeth = 0.0052 inch per tooth Next:  Review Questions for Chapter 3 Back to Contents Page Updated Jan. 9, 2002 Copyright © 1988-2002 by George Stanton and Bill Hemphill All Rights Reserved

CATEGORY | ROUTER BITS | Solid Surface Router Bits (for CORIAN)

The horsepower required to run the spindle is the product of the Material Removal Rate (MRR) * the Material Machining Factor (MMF). The MMF, a function of the machineability rating of the workpiece material, is shown in Table 3.3. Table 3.3 Material Machining Factors (MMF) Material MMF Aluminum0.12 Brass0.4 Low-carbon steel1.25 High-carbon & low alloy steel1.5 - 1.9 Stainless and tool steels1.8 - 2.2 Superalloys2.5 - 5.0 More Mathematics For example, suppose a 1-inch-diameter end mill was making a 1-inch-deep cut in low-carbon steel with a feedrate of 10 inches per minute (IPM). The MRR is 1 (depth) * 1 (width) * 10 (IPM) = 10. Table 3.3 shows low-carbon steel has an MMF of 1.25. Thus, it would require 10 (MRR) * 1.25 (MMF) = 12.5 HP to drive the spindle. If the spindle was driven with a 5 HP motor, it could stall and perhaps break the cutter. The feedrate (or depth of cut) would have to be reduced to lower the MRR, so a 5 HP or lower load was placed on the spindle. Since: HP = MRR * MMF and MRR = depth of cut * width of cut * feedrate it follows that: HP = MMF * depth of cut * width of cut * feedrate. Thus, Feedrate = HP --------------------------------------- MMF * depth of cut * width of cut In the previous example, if the spindle motor were 5 HP Feedrate = 5 HP --------------------------------------- 1.25 (MMF) * 1 (depth) * 1 (width) = 4 IPM   Per tooth chip load = feedrate (IPM) --------------------------------------- spindle RPM * cutter teeth Assuming a spindle speed of 400 RPM and a 2-flute cutter, Chip load = 4.0 IPM --------------------------------------- 400 RPM * 2 teeth = 0.005 inch per tooth Suppose in the previous example the workpiece material is changed to aluminum (with an MMF of 0.12). Spindle horsepower is 5. Spindle RPM is 4000. Feedrate = 5 HP --------------------------------------- 0.12 (MMF) * 1 (depth) * 1 (width) = 41.7 IPM   Chip load = 41.7 IPM --------------------------------------- 4000 RPM * 2 teeth = 0.0052 inch per tooth Next:  Review Questions for Chapter 3 Back to Contents Page Updated Jan. 9, 2002 Copyright © 1988-2002 by George Stanton and Bill Hemphill All Rights Reserved

it follows that: HP = MMF * depth of cut * width of cut * feedrate. Thus, Feedrate = HP --------------------------------------- MMF * depth of cut * width of cut In the previous example, if the spindle motor were 5 HP Feedrate = 5 HP --------------------------------------- 1.25 (MMF) * 1 (depth) * 1 (width) = 4 IPM   Per tooth chip load = feedrate (IPM) --------------------------------------- spindle RPM * cutter teeth Assuming a spindle speed of 400 RPM and a 2-flute cutter, Chip load = 4.0 IPM --------------------------------------- 400 RPM * 2 teeth = 0.005 inch per tooth Suppose in the previous example the workpiece material is changed to aluminum (with an MMF of 0.12). Spindle horsepower is 5. Spindle RPM is 4000. Feedrate = 5 HP --------------------------------------- 0.12 (MMF) * 1 (depth) * 1 (width) = 41.7 IPM   Chip load = 41.7 IPM --------------------------------------- 4000 RPM * 2 teeth = 0.0052 inch per tooth Next:  Review Questions for Chapter 3 Back to Contents Page Updated Jan. 9, 2002 Copyright © 1988-2002 by George Stanton and Bill Hemphill All Rights Reserved

For drilling, the MRR is the product of the area of a circle the diameter of the drill (drill diameter squared * pi/4) * the feedrate in inches per revolution * the spindle speed in RPM. The horsepower required to run the spindle is the product of the Material Removal Rate (MRR) * the Material Machining Factor (MMF). The MMF, a function of the machineability rating of the workpiece material, is shown in Table 3.3. Table 3.3 Material Machining Factors (MMF) Material MMF Aluminum0.12 Brass0.4 Low-carbon steel1.25 High-carbon & low alloy steel1.5 - 1.9 Stainless and tool steels1.8 - 2.2 Superalloys2.5 - 5.0 More Mathematics For example, suppose a 1-inch-diameter end mill was making a 1-inch-deep cut in low-carbon steel with a feedrate of 10 inches per minute (IPM). The MRR is 1 (depth) * 1 (width) * 10 (IPM) = 10. Table 3.3 shows low-carbon steel has an MMF of 1.25. Thus, it would require 10 (MRR) * 1.25 (MMF) = 12.5 HP to drive the spindle. If the spindle was driven with a 5 HP motor, it could stall and perhaps break the cutter. The feedrate (or depth of cut) would have to be reduced to lower the MRR, so a 5 HP or lower load was placed on the spindle. Since: HP = MRR * MMF and MRR = depth of cut * width of cut * feedrate it follows that: HP = MMF * depth of cut * width of cut * feedrate. Thus, Feedrate = HP --------------------------------------- MMF * depth of cut * width of cut In the previous example, if the spindle motor were 5 HP Feedrate = 5 HP --------------------------------------- 1.25 (MMF) * 1 (depth) * 1 (width) = 4 IPM   Per tooth chip load = feedrate (IPM) --------------------------------------- spindle RPM * cutter teeth Assuming a spindle speed of 400 RPM and a 2-flute cutter, Chip load = 4.0 IPM --------------------------------------- 400 RPM * 2 teeth = 0.005 inch per tooth Suppose in the previous example the workpiece material is changed to aluminum (with an MMF of 0.12). Spindle horsepower is 5. Spindle RPM is 4000. Feedrate = 5 HP --------------------------------------- 0.12 (MMF) * 1 (depth) * 1 (width) = 41.7 IPM   Chip load = 41.7 IPM --------------------------------------- 4000 RPM * 2 teeth = 0.0052 inch per tooth Next:  Review Questions for Chapter 3 Back to Contents Page Updated Jan. 9, 2002 Copyright © 1988-2002 by George Stanton and Bill Hemphill All Rights Reserved

For milling, the material removal rate (MRR) can be calculated as follows: MMR = depth of cut * the width of cut * spindle speed * the feed per tooth (chip load) * the number of teeth * the spindle speed (RPM). For lathe cuts, the MRR is the product of the mean workpiece circumference * the depth of cut * feedrate (in inches per revolution) * the spindle speed. The mean workpiece circumference is the circumference halfway down the depth of the cut. For drilling, the MRR is the product of the area of a circle the diameter of the drill (drill diameter squared * pi/4) * the feedrate in inches per revolution * the spindle speed in RPM. The horsepower required to run the spindle is the product of the Material Removal Rate (MRR) * the Material Machining Factor (MMF). The MMF, a function of the machineability rating of the workpiece material, is shown in Table 3.3. Table 3.3 Material Machining Factors (MMF) Material MMF Aluminum0.12 Brass0.4 Low-carbon steel1.25 High-carbon & low alloy steel1.5 - 1.9 Stainless and tool steels1.8 - 2.2 Superalloys2.5 - 5.0 More Mathematics For example, suppose a 1-inch-diameter end mill was making a 1-inch-deep cut in low-carbon steel with a feedrate of 10 inches per minute (IPM). The MRR is 1 (depth) * 1 (width) * 10 (IPM) = 10. Table 3.3 shows low-carbon steel has an MMF of 1.25. Thus, it would require 10 (MRR) * 1.25 (MMF) = 12.5 HP to drive the spindle. If the spindle was driven with a 5 HP motor, it could stall and perhaps break the cutter. The feedrate (or depth of cut) would have to be reduced to lower the MRR, so a 5 HP or lower load was placed on the spindle. Since: HP = MRR * MMF and MRR = depth of cut * width of cut * feedrate it follows that: HP = MMF * depth of cut * width of cut * feedrate. Thus, Feedrate = HP --------------------------------------- MMF * depth of cut * width of cut In the previous example, if the spindle motor were 5 HP Feedrate = 5 HP --------------------------------------- 1.25 (MMF) * 1 (depth) * 1 (width) = 4 IPM   Per tooth chip load = feedrate (IPM) --------------------------------------- spindle RPM * cutter teeth Assuming a spindle speed of 400 RPM and a 2-flute cutter, Chip load = 4.0 IPM --------------------------------------- 400 RPM * 2 teeth = 0.005 inch per tooth Suppose in the previous example the workpiece material is changed to aluminum (with an MMF of 0.12). Spindle horsepower is 5. Spindle RPM is 4000. Feedrate = 5 HP --------------------------------------- 0.12 (MMF) * 1 (depth) * 1 (width) = 41.7 IPM   Chip load = 41.7 IPM --------------------------------------- 4000 RPM * 2 teeth = 0.0052 inch per tooth Next:  Review Questions for Chapter 3 Back to Contents Page Updated Jan. 9, 2002 Copyright © 1988-2002 by George Stanton and Bill Hemphill All Rights Reserved

We have over 70 years of experience designing and fabricating chip conveyors and material handling equipment for machine tool, metal working and material handling industries. Each of our conveyors are durable, high-quality, long-lasting and can be custom-built to meet your needs.

Having the correct metal chip conveyor for your application can have a huge impact on your business. We offer a variety of filtration and chip processing options. These can be combined into a complete solution to meet the requirements of the application.

The classic method of determining the feedrate is to decide upon a chip load (how thick a chip you wish each tooth to peel off), typically from 0.003 to 0.020 inch. Then calculate what the spindle speed should be. Next, count the number of teeth on the cutter. Then multiply the three together: Feedrate (IPM) = chipload (IN) * RPM * teeth The results of that equation, while adequate for most manual machining operations, is often inadequate for high spindle horsepower N/C machines. In addition to being uneconomical, a feedrate that is too slow can hasten cutting tool failure. As each tooth peels off a chip, it also work hardens the workpiece surface. The depth of this workhardened layer can range from a few millionths of an inch to over a thousandth of an inch. If the feedrate is too slow, the cutting edge of each tooth can be cutting through the workhardened layer (dulling the cutting edge) instead of slicing beneath the layer. A feedrate that is too high can stall the spindle or break the cutter--or both. The feedrate that is required to stall the spindle depends upon the following factors: Horsepower of the spindle drive motor, Material Machining Factor (MMF), the machineability of the workpiece material, Material Removal Rate (MRR), the volume (cubic inches) of material being removed per minute. MRR, in turn, is dependent on the depth of cut, diameter of the cutter, and feedrate. Geometry of the cutter, Sharpness of the cutter. The geometry of the cutter is a minor factor and can be ignored. Good machining practice presupposes the cutter is sharp, so that factor can be ignored. Material Removal Rate (MRR) For milling, the material removal rate (MRR) can be calculated as follows: MMR = depth of cut * the width of cut * spindle speed * the feed per tooth (chip load) * the number of teeth * the spindle speed (RPM). For lathe cuts, the MRR is the product of the mean workpiece circumference * the depth of cut * feedrate (in inches per revolution) * the spindle speed. The mean workpiece circumference is the circumference halfway down the depth of the cut. For drilling, the MRR is the product of the area of a circle the diameter of the drill (drill diameter squared * pi/4) * the feedrate in inches per revolution * the spindle speed in RPM. The horsepower required to run the spindle is the product of the Material Removal Rate (MRR) * the Material Machining Factor (MMF). The MMF, a function of the machineability rating of the workpiece material, is shown in Table 3.3. Table 3.3 Material Machining Factors (MMF) Material MMF Aluminum0.12 Brass0.4 Low-carbon steel1.25 High-carbon & low alloy steel1.5 - 1.9 Stainless and tool steels1.8 - 2.2 Superalloys2.5 - 5.0 More Mathematics For example, suppose a 1-inch-diameter end mill was making a 1-inch-deep cut in low-carbon steel with a feedrate of 10 inches per minute (IPM). The MRR is 1 (depth) * 1 (width) * 10 (IPM) = 10. Table 3.3 shows low-carbon steel has an MMF of 1.25. Thus, it would require 10 (MRR) * 1.25 (MMF) = 12.5 HP to drive the spindle. If the spindle was driven with a 5 HP motor, it could stall and perhaps break the cutter. The feedrate (or depth of cut) would have to be reduced to lower the MRR, so a 5 HP or lower load was placed on the spindle. Since: HP = MRR * MMF and MRR = depth of cut * width of cut * feedrate it follows that: HP = MMF * depth of cut * width of cut * feedrate. Thus, Feedrate = HP --------------------------------------- MMF * depth of cut * width of cut In the previous example, if the spindle motor were 5 HP Feedrate = 5 HP --------------------------------------- 1.25 (MMF) * 1 (depth) * 1 (width) = 4 IPM   Per tooth chip load = feedrate (IPM) --------------------------------------- spindle RPM * cutter teeth Assuming a spindle speed of 400 RPM and a 2-flute cutter, Chip load = 4.0 IPM --------------------------------------- 400 RPM * 2 teeth = 0.005 inch per tooth Suppose in the previous example the workpiece material is changed to aluminum (with an MMF of 0.12). Spindle horsepower is 5. Spindle RPM is 4000. Feedrate = 5 HP --------------------------------------- 0.12 (MMF) * 1 (depth) * 1 (width) = 41.7 IPM   Chip load = 41.7 IPM --------------------------------------- 4000 RPM * 2 teeth = 0.0052 inch per tooth Next:  Review Questions for Chapter 3 Back to Contents Page Updated Jan. 9, 2002 Copyright © 1988-2002 by George Stanton and Bill Hemphill All Rights Reserved

Jorgensen excels at providing innovative, often custom, chip conveyors and chip processing systems. As a result, Jorgensen is the preferred source of metal chip conveyors for many industry leading machine tool manufacturers and distribution centers.

Per tooth chip load = feedrate (IPM) --------------------------------------- spindle RPM * cutter teeth Assuming a spindle speed of 400 RPM and a 2-flute cutter, Chip load = 4.0 IPM --------------------------------------- 400 RPM * 2 teeth = 0.005 inch per tooth Suppose in the previous example the workpiece material is changed to aluminum (with an MMF of 0.12). Spindle horsepower is 5. Spindle RPM is 4000. Feedrate = 5 HP --------------------------------------- 0.12 (MMF) * 1 (depth) * 1 (width) = 41.7 IPM   Chip load = 41.7 IPM --------------------------------------- 4000 RPM * 2 teeth = 0.0052 inch per tooth Next:  Review Questions for Chapter 3 Back to Contents Page Updated Jan. 9, 2002 Copyright © 1988-2002 by George Stanton and Bill Hemphill All Rights Reserved

Feedrate (IPM) = chipload (IN) * RPM * teeth The results of that equation, while adequate for most manual machining operations, is often inadequate for high spindle horsepower N/C machines. In addition to being uneconomical, a feedrate that is too slow can hasten cutting tool failure. As each tooth peels off a chip, it also work hardens the workpiece surface. The depth of this workhardened layer can range from a few millionths of an inch to over a thousandth of an inch. If the feedrate is too slow, the cutting edge of each tooth can be cutting through the workhardened layer (dulling the cutting edge) instead of slicing beneath the layer. A feedrate that is too high can stall the spindle or break the cutter--or both. The feedrate that is required to stall the spindle depends upon the following factors: Horsepower of the spindle drive motor, Material Machining Factor (MMF), the machineability of the workpiece material, Material Removal Rate (MRR), the volume (cubic inches) of material being removed per minute. MRR, in turn, is dependent on the depth of cut, diameter of the cutter, and feedrate. Geometry of the cutter, Sharpness of the cutter. The geometry of the cutter is a minor factor and can be ignored. Good machining practice presupposes the cutter is sharp, so that factor can be ignored. Material Removal Rate (MRR) For milling, the material removal rate (MRR) can be calculated as follows: MMR = depth of cut * the width of cut * spindle speed * the feed per tooth (chip load) * the number of teeth * the spindle speed (RPM). For lathe cuts, the MRR is the product of the mean workpiece circumference * the depth of cut * feedrate (in inches per revolution) * the spindle speed. The mean workpiece circumference is the circumference halfway down the depth of the cut. For drilling, the MRR is the product of the area of a circle the diameter of the drill (drill diameter squared * pi/4) * the feedrate in inches per revolution * the spindle speed in RPM. The horsepower required to run the spindle is the product of the Material Removal Rate (MRR) * the Material Machining Factor (MMF). The MMF, a function of the machineability rating of the workpiece material, is shown in Table 3.3. Table 3.3 Material Machining Factors (MMF) Material MMF Aluminum0.12 Brass0.4 Low-carbon steel1.25 High-carbon & low alloy steel1.5 - 1.9 Stainless and tool steels1.8 - 2.2 Superalloys2.5 - 5.0 More Mathematics For example, suppose a 1-inch-diameter end mill was making a 1-inch-deep cut in low-carbon steel with a feedrate of 10 inches per minute (IPM). The MRR is 1 (depth) * 1 (width) * 10 (IPM) = 10. Table 3.3 shows low-carbon steel has an MMF of 1.25. Thus, it would require 10 (MRR) * 1.25 (MMF) = 12.5 HP to drive the spindle. If the spindle was driven with a 5 HP motor, it could stall and perhaps break the cutter. The feedrate (or depth of cut) would have to be reduced to lower the MRR, so a 5 HP or lower load was placed on the spindle. Since: HP = MRR * MMF and MRR = depth of cut * width of cut * feedrate it follows that: HP = MMF * depth of cut * width of cut * feedrate. Thus, Feedrate = HP --------------------------------------- MMF * depth of cut * width of cut In the previous example, if the spindle motor were 5 HP Feedrate = 5 HP --------------------------------------- 1.25 (MMF) * 1 (depth) * 1 (width) = 4 IPM   Per tooth chip load = feedrate (IPM) --------------------------------------- spindle RPM * cutter teeth Assuming a spindle speed of 400 RPM and a 2-flute cutter, Chip load = 4.0 IPM --------------------------------------- 400 RPM * 2 teeth = 0.005 inch per tooth Suppose in the previous example the workpiece material is changed to aluminum (with an MMF of 0.12). Spindle horsepower is 5. Spindle RPM is 4000. Feedrate = 5 HP --------------------------------------- 0.12 (MMF) * 1 (depth) * 1 (width) = 41.7 IPM   Chip load = 41.7 IPM --------------------------------------- 4000 RPM * 2 teeth = 0.0052 inch per tooth Next:  Review Questions for Chapter 3 Back to Contents Page Updated Jan. 9, 2002 Copyright © 1988-2002 by George Stanton and Bill Hemphill All Rights Reserved

Jorgensen Conveyor and Filtration Solutions - An Innovance Company - 2024 - All Rights Reserved (262) 242-3089 - Privacy Policy